A hollow spherical shell at outer radius R floats just submerged under the water surface.

Question:

A hollow spherical shell at outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$. If the specific gravity of the shell material

is $\frac{27}{8}$ w.r.t water, the value of $r$ is :

  1. (1) $\frac{8}{9} R$

  2. (2) $\frac{4}{9} R$

  3. (3) $\frac{2}{3} R$

  4. (4) $\frac{1}{3} R$


Correct Option: 1

Solution:

(1) In equilibrium, $m g=F_{e}$

$F_{B}=V \rho_{0} g$ and mass $=$ volume $\times$ density

$\frac{4}{3} \pi\left(R^{3}-r^{3}\right) \rho_{0} g=\frac{4}{3} \pi R^{3} \rho_{w} g$

Given, relative density, $\frac{\rho_{0}}{\rho_{w}}=\frac{27}{8}$

$\Rightarrow\left[1-\left(\frac{r}{R}\right)^{3}\right] \frac{27}{8} \rho_{w}=\rho_{w}$

$\Rightarrow 1-\frac{r^{3}}{R^{3}}=\frac{9}{27} \Rightarrow 1-\frac{1}{3}=\frac{r^{3}}{R^{3}} \Rightarrow \frac{2}{3}=\frac{r^{3}}{R^{3}}$

$\Rightarrow \frac{r}{R}=\left(\frac{2}{3}\right)^{1 / 3} \Rightarrow 1-\frac{r^{3}}{R^{3}}=\frac{8}{27}$

$\Rightarrow \frac{r^{3}}{R^{3}}=1-\frac{8}{27}=\frac{19}{27}$

$\therefore r=0.89 R=\frac{8}{9} R$

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